The speed of sound in a solid would be 6000 metres per second
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:

Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation.

is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.

Our final equation is:

Colatitude is:

The answer is:
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of

from the center of the pattern. In the formula, m is the order of the minimum,

the wavelenght,

the distance of the screen from the slit and

the width of the slit.
In our problem, the distance of the first-order band (m=1) is

. The distance of the screen is D=86 cm while the wavelength is

. Using these data and re-arranging the formula, we can find a, the width of the slit:
Answer:
O²⁻
Explanation:
Number of protons = 8
Number of neutrons = 9
Number of electrons = 10
What type of atom or ion is it = ?
Solution:
Protons are the positively charged particle in an atom
Neutrons do not carry any charges
Electrons are negatively charged particles
For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.
Now,
Charge = Number of protons - Number of electrons
Charge = 8 - 10 = -2
The charge on the atom is -2 and so it is an oxygen ion with -2 charge
The ion is O²⁻
Answer:
The mass of the mud is 3040000 kg.
Explanation:
Given that,
length = 2.5 km
Width = 0.80 km
Height = 2.0 m
Length of valley = 0.40 km
Width of valley = 0.40 km
Density = 1900 Kg/m³
Area = 4.0 m²
We need to calculate the mass of the mud
Using formula of density


Where, V = volume of mud
= density of mud
Put the value into the formula


Hence, The mass of the mud is 3040000 kg.